Question

Let R be the region bounded by

y
=
7
sin
(
π
2
x
)
,
y
=
7
(
x
−
2
)
2
, and
y
=
x
+
6
, and containing the point (2,7).

Answer 1

a. The area of
`R` is given by the integral

`\displaystyle \int_1^2 (x + 6) - 7\sin\left(\frac{\pi x}2\right) \, dx + \int_2^(22/7) (x+6) - 7(x-2)^2 \, dx \approx 9.36`

b. Use the shell method. Revolving
`R` about the
`x`-axis generates shells with height
`h=x+6-7\sin\left(\frac{\pi x}2\right)` when
`1\le x\le 2`, and
`h=x+6-7(x-2)^2` when
`2\le x\le\frac{22}7`. With radius
`r=x`, each shell of thickness
`\Delta x` contributes a volume of
`2\pi r h \Delta x`, so that as the number of shells gets larger and their thickness gets smaller, the total sum of their volumes converges to the definite integral

`\displaystyle 2\pi \int_1^2 x \left((x + 6) - 7\sin\left(\frac{\pi x}2\right)\right) \, dx + 2\pi \int_2^(22/7) x\left((x+6) - 7(x-2)^2\right) \, dx \approx 129.56`

c. Use the washer method. Revolving
`R` about the
`y`-axis generates washers with outer radius
`r_(\rm out) = x+6`, and inner radius
`r_(\rm in)=7\sin\left(\frac{\pi x}2\right)` if
`1\le x\le2` or
`r_(\rm in) = 7(x-2)^2` if
`2\le x\le\frac{22}7`. With thickness
`\Delta x`, each washer has volume
`\pi (r_(\rm out)^2 - r_(\rm in)^2) \Delta x`. As more and thinner washers get involved, the total volume converges to

`\displaystyle \pi \int_1^2 (x+6)^2 - \left(7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \pi \int_2^(22/7) (x+6)^2 - \left(7(x-2)^2\right)^2 \, dx \approx 304.16`

d. The side length of each square cross section is
`s=x+6 - 7\sin\left(\frac{\pi x}2\right)` when
`1\le x\le2`, and
`s=x+6-7(x-2)^2` when
`2\le x\le\frac{22}7`. With thickness
`\Delta x`, each cross section contributes a volume of
`s^2 \Delta x`. More and thinner sections lead to a total volume of

`\displaystyle \int_1^2 \left(x+6-7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \int_2^(22/7) \left(x+6-7(x-2)^2\right) ^2\, dx \approx 56.70`